Step 1:

Let $X$ denote the number of doublets.

Possible doublets are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$

Clearly $X$ can take values as $0,1,2,3$ and 4

Probability of getting a doublet =$\large\frac{6}{36}=\frac{1}{6}$

Probability of not getting a doublet =$1-\large\frac{1}{36}=\frac{1}{6}$

Step 2:

Now $P(X=0)=P$(no doublets)

$\qquad\qquad\quad\;\;=\large\frac{5}{6}\times \frac{5}{6}\times\frac{5}{6}\times \frac{5}{6}$

$\qquad\qquad\quad\;\;=\large\frac{625}{1296}$

Step 3:

$P(X=1)=P$(one doublet and three non-doublets)

$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}$

$\qquad\quad\;\;\;=4\big(\large\frac{1}{6}\times \frac{5^3}{6^3}\big)$

$\qquad\quad\;\;\;=\large\frac{500}{1296}$

$\qquad\quad\;\;\;=\large\frac{125}{324}$

Step 4:

$P(X=2)=P$(two doublet and two non-doublets)

$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{1}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}$

$\qquad\quad\;\;\;=4\big(\large\frac{1}{6^2}\times \frac{5^2}{6^2}\big)$

$\qquad\quad\;\;\;=\large\frac{100}{1296}$

$\qquad\quad\;\;\;=\large\frac{25}{324}$

Step 5:

$P(X=3)=P$(three doublets and one non-doublets)

$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}+\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}$

$\qquad\quad\;\;\;=4\big(\large\frac{1}{6^3}\times \frac{5}{6}\big)$

$\qquad\quad\;\;\;=\large\frac{20}{1296}$

$\qquad\quad\;\;\;=\large\frac{5}{324}$

Step 6:

$P(X=4)=\large\frac{1}{6}\times \large\frac{1}{6}\times\large\frac{1}{6}\times\large\frac{1}{6}$

$\qquad\quad\;\;\;=\large\frac{1}{1296}$

Thus the required probability distribution is